1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Convergence of {fn} wrt to C(X) metric

  1. Jan 24, 2012 #1
    Let C(X) be the set of continuous complex-valued bounded functions with domain X. Let {fn} be a sequence of functions on C. We say, that fn converges to f wrt to the metric of C(X) iff fn converges to f uniformly.

    By definition of the uniform convergence, for any ε>0 there exists integer N s.t. n ≥ N implies |fn(x) - f(x)| ≤ ε for every x in X. Hence by the definition of supremum norm, the inequality is equivalent to sup(fn(x) - f(x)) ≤ ε.

    I conclude, that:
    1) the uniform convergence imposes that for every ε>0, there exists N from which on the functions fn and f are within this ε on whole X. Hence it always implies convergence wrt the metric of C(X), since exactly this ε can be used as the supremum of the distance fn and f.
    2) What about the converse? I think, that if {fn} converges to f wrt to the metric of C(X), then the existence of such ε>0, being the supremum of the distance on whole X, implies the uniform convergence, as this ε can be used in its definition and the existence of N is guaranteed.

    Am I right or not?
     
  2. jcsd
  3. Jan 24, 2012 #2

    Fredrik

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Is the question if it's true that ##f_n\to f## uniformly if and only if ##f_n\to f## with respect to the metric defined from the norm defined by ##\|f\|=\sup_{x\in X}|f(x)|## for all f in C(X)? If that's what you're asking, the answer is yes. This is how I would start a proof of that theorem:

    Let ##\varepsilon>0## be arbitrary.

    Suppose that ##f_n\to f## with respect to the norm (metric). Choose N such that for all n≥N,
    $$\|f_n-f\|<\varepsilon.$$ This choice ensures that for all n≥N and all x in X,
    $$|f_n(x)-f(x)| \dots $$
    So ##f_n\to f## uniformly.

    Can you fill in the details (at the dots)? The proof of the converse is very similar.
     
  4. Jan 24, 2012 #3
    Thank you, Frederic. This was exactly what I was thinking about, and, roughly said, your answer follows the idea provided by me in (2). The goal was to ensure that I'm not missing something.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook